We will learn how to find the intercepts on the axes made by a circle.
The lengths of intercepts made by the circle x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 with X and Y axes are 2\(\mathrm{\sqrt{g^{2}  c}}\) and 2\(\mathrm{\sqrt{f^{2}  c}}\) respectively.
Proof:
Let the given equation of the circle be x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 ………. (1)
Clearly, the centre of the circle is c (g, f) and the radius = \(\mathrm{\sqrt{g^{2} + f^{2} c}}\)
Let AB be the intercept made by the given circle on xaxe.
Since on xaxis, y = 0. Therefore, xcoordinates of the points A and B are the
roots of the equation x\(^{2}\) + 2gx + c = 0.
Let x\(_{1}\) and x\(_{2}\) be the xcoordinates of the points A and B respectively. Then, x\(_{1}\) and x\(_{2}\) also the roots of the equation x\(^{2}\) + 2gx + c = 0.
Therefore, x\(_{1}\) + x\(_{2}\) =  2g and x\(_{1}\)x\(_{2}\) = c
Clearly the intercept on xaxis = AB
= x\(_{2}\)  x\(_{1}\) = \(\mathrm{\sqrt{(x_{2}  x_{1})^{2}}}\)
= \(\mathrm{\sqrt{(x_{2} + x_{1})^{2}  4x_{1}x_{2}}}\)
= \(\mathrm{\sqrt{4g^{2}  4c}}\)
= 2\(\mathrm{\sqrt{g^{2}  c}}\)
Therefore, the intercept made by the circle (1) on the xaxis = 2\(\mathrm{\sqrt{g^{2}  c}}\)
Again,
Let DE be the intercept made by the given circle on yaxe. Since on yaxis, x = 0. Therefore, ycoordinates of the points D and E are the roots of the equation y\(^{2}\) + 2fy + c = 0.
Let y\(_{1}\) and y\(_{2}\) be the xcoordinates of the points D and E respectively. Then, y\(_{1}\) and y\(_{2}\) also the roots of the equation y\(^{2}\) + 2fy + c = 0
Therefore, y\(_{1}\) + y\(_{2}\) =  2f and y\(_{1}\)y\(_{2}\) = c
Clearly the intercept on yaxis = DE
= y\(_{2}\)  y\(_{1}\) = \(\mathrm{\sqrt{(y_{2}  y_{1})^{2}}}\)
= \(\mathrm{\sqrt{(y_{2} + y_{1})^{2} – 4y_{1}y_{2}}}\)
= \(\mathrm{\sqrt{4f^{2}  4c}}\)
= 2\(\mathrm{\sqrt{f^{2}  c}}\)
Therefore, the intercept made by the circle (1) on the yaxis = 2\(\mathrm{\sqrt{f^{2}  c}}\)
`Solved examples to find the intercepts made by a given circle on the coordinate axes:
1. Find the length of the xintercept and yintercept made by the circle x\(^{2}\) + y\(^{2}\)  4x 6y  5 = 0 with the coordinate axes.
Solution:
Given equation of the circle is x\(^{2}\) + y\(^{2}\)  4x 6y  5 = 0.
Now comparing the given equation with the general equation of the circle x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0, we get g = 2 and f = 3 and c = 5
Therefore, length of the xintercept = 2\(\mathrm{\sqrt{g^{2}  c}}\) = 2\(\mathrm{\sqrt{4  (5) }}\) = 2√9 = 6.
The length of the yintercept = 2\(\mathrm{\sqrt{f^{2}  c}}\) = 2\(\mathrm{\sqrt{9  (5) }}\) = 2√14.
2. Find the equation of a circle which touches the yaxis at a distance 3 from the origin and cuts an intercept of 8 units with the positive direction of xaxis.
Solution:
Let the equation of the circle be x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 …………….. (i)
According to the problem, the equation (i) touches the yaxis
Therefore, c = f\(^{2}\) ………………… (ii)
Again, the point (0, 3) lies on the circle (i).
Therefore, putting the value of x = 0 and y = 3 in (i) we get,
9  6f + c = 0 …………………… (iii)
From (ii) and (iii), we get 9  6f + f\(^{2}\) = 0 ⇒ (f  3)\(^{2}\) = 0 ⇒ f  3 = 0 ⇒ f = 3
Now putting f = 3 in (i) we get, c = 9
Again, according to the problem the equation of the circle (i) cuts an intercept of 8 units with the positive direction of xaxis.
Therefore,
2\(\mathrm{\sqrt{g^{2}  c}}\) = 8
⇒ 2\(\mathrm{\sqrt{g^{2}  9}}\) = 8
⇒ \(\mathrm{\sqrt{g^{2}  9}}\) = 4
⇒ g\(^{2}\)  9 = 16, [Squaring both sides]
⇒ g\(^{2}\) = 16 + 9
⇒ g\(^{2}\) = 25
⇒ g = ±5.
Hence, the required equation of the circle is x^2 + y^2 ± 10x + 6y + 9 = 0.
`● The Circle
11 and 12 Grade Math
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